# Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

## Theorem

Let $\struct {R, +, \circ}$ be a commutative ring with unity.

Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$.

Let $X \in R$ be transcendental over $D$

Then the ring of polynomials $D \sqbrk X$ in $X$ over $D$ exists.

## Proof

Suppose that $D \sqbrk X$ exists.

Let $\ds \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$.

Then $\map P X$ corresponds to, and is completely described by, the ordered tuple of coefficients $\tuple {a_0, a_1, \dotsc, a_n, 0_D, 0_D, 0_D, \dotsc}$.

Consider the set $S$ of infinite sequences of elements of $D$ which are eventually $0_D$.

That is, whose elements are of the form $\tuple {b_0, b_1, \dotsc, b_n, 0_D, 0_D, 0_D, \dotsc}$ where $b_0, \ldots, b_n \in D$.

Consider the polynomial ring over $S$ by defining the operations:

 $(1)$ $:$ Ring Addition: $\ds \sequence {r_0, r_1, r_2, \ldots} + \sequence {s_0, s_1, s_2, \ldots}$ $\ds =$ $\ds \sequence {r_0 + s_0, r_1 + s_1, r_2 + s_2, \ldots}$ $(2)$ $:$ Ring Negative: $\ds -\sequence {r_0, r_1, r_2, \ldots}$ $\ds =$ $\ds \sequence {-r_0, -r_1, -r_2, \ldots}$ $(3)$ $:$ Ring Product: $\ds \sequence {r_0, r_1, r_2, \ldots} \circ \sequence {s_0, s_1, s_2, \ldots}$ $\ds =$ $\ds \sequence {t_0, t_1, t_2, \ldots}$ where $\ds t_i = \sum_{j \mathop + k \mathop = i} r_j s_k$

From Polynomial Ring of Sequences is Ring we have that $\struct {S, +, \circ}$ is a ring.